# Physical Sciences Grade 12 P2 Limpopo November 2014 Memo

> Grade 12 Physical Sciences Paper 2 memorandum from the Limpopo November 2014, 10 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/ldoe-physc-p2-sept-p2-2014-memo/55580970-7b00-43de-ad28-f0b6ddbf91f2
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 12
- Document type: Memorandum
- Year: 2014
- Exam period: Limpopo November
- Paper: 2
- Curriculum: CAPS
- Pages: 10

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Physical Sciences · Grade 12 · Limpopo November · 2014. Memorandum, 10 pages. Read online or download the PDF.

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### Page 1

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                                  NATIONAL
                              SENIOR CERTIFICATE

                                    GRADE 12

                     PHYSICAL SCIENCES: CHEMISTRY (P2)
                     FISIESE WETENSKAPPE: CHEMIE (Vr2)
                              SEPTEMBER 2014

                                MEMORANDUM

MARKS: 150

                       This memorandum consists of 10 pages.

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Physical Science/P2 2 LimpopoDoE/September 2014
NCS-Memorandum

QUESTION/ VRAAG 1

11. Cvv (2)
12 Dvv (2)
1.3. Cvv (2)
14. Avy (2)
15 Avy (2)
16  Cvv (2)
1.7 Byy (2)
18 Byy (2)
19 Avy (2)
1.10 Dvv (2)
[20]
QUESTION/VRAAG 2
2.1
2.1.1 Uv (1)
2.1.2 Carboxyl groupY / karboksielgroep (1)
2.1.3 Aldehyde / alkanal v//aldehied (1)
2.1.4 2-bromo-1,4-dichloroY butane ¥ /2-bromo-1,4-dichworoburaan
of 2-broom-1,4-dichloorobutaan (2)
2.1.5 polythenev /politeen
ACCEPT: Poly(ethene) (1)
2.1.6. 2CgHiqg + 1902” > 12CO02+14H20Y = Balv
ReactantsY ProductsY BalancingY (3)
2.1.7. 4-Methyl” hexan-1-ol” / 4-métielheksan-1-ol Y
Accept: 4-Methyl-1-hexanol (2)
2.1.8. CnH2nOo” (1)
2.2 The reddish-brown (bromine water) solution decolourises/
The reddish-brown colour disappears” // die rooi-bruin kleur
verdwyn/oplossing ontkleur (1)
2.3
H H HoH
‘cmc’ 4 Br-Br ——> H~C—CNH
/ \ | |
H H Br sor

Notes

v alkene/alkene =v Bromine/ broom

Y functional group correct (product)/funksionele group van produk
Y whole structure correct (product)/ hele struktuur van produk

Do not penalise for Bro

(4)
[17]

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Physical Science/P2                       3                LimpopoDoE/September 2014
                                    NCS-Memorandum
QUESTION/VRAAG 3

3.1.   A series of organic compounds that can be described by the same general
       formula /’n reeks organiese verbindings wat dieselfde algemene formule het
       A series of organic compounds whose successive members differ by a CH2 unit
       only/ ’n reeks organiese verbindings wat met ‘n CH2 eenheid van mekaar verskil
       A series of organic compounds having the same functional group./ / ’n reeks
       organiese verbindings watdieselfde funksionele groep het                  (2)
3.2.   Chloromethane / chloormetaan/chlorometaan
          Fluoromethane /fluoormetaan/fluorometaan                              (2)
3.3.   Boiling points increase as the mass/size of the molecule increases.
OR     Boiling points decrease as the mass/size of the molecule decreases /
       kookpunte neem toe soos die massa/grootte van die molekules toeneem OF
       kookpunte neem af soos die massa/grootte van die molekules afneem (2)
3.4.   As the molecular mass increases, the strength of Van Der Waals forces
       (IMFs) increases. Hence more energy is needed to overcome the
       Van Der Waals forces (IMFs). Thus the boiling points increase./
       Soos die molekulêre massa toeneem word die Van der Waalskragte sterker.
       Dus word meer energie benodig om die Van der Waalskragte (IMK) te oorkom,
       dus neem die kookpunte toe                                         (3)
3.5.

            Criteria for investigative question/ kriteria                   Mark
            Refers to relationship between dependent and independent        
            variables/ verwys na verhouding tussen die afhanklike en
            onafhanklike veranderlikes
            It is a question – not stated as an aim/ is ‘n vraag            

         For example/ byvoorbeeld:
         What is the relationship between boiling point and type of organic
         compound/ functional group/ homologous series? // wat is die verhouding
         tussen kookpunt en die soort organiese verbinding/funksionele groep/
         homoloë reeks?                                                      (2)
3.6 Molecular mass  /molekulêre massa                                         (1)
3.7 The carboxylic acids have higher boiling points than alcohols
   which have higher boiling points than the aldehydes with the same molecular
   mass. / die karboksielsure het ‘n hoer kookpunt as die alkohole en hulle het ‘n
   hoer kookpunt as die aldehiedes                                             (2)
                                                                                [14]
QUESTION/VRAAG 4

4.1 Cracking  /Dehydrogenation/ elimination//kraking/eliminasie/dehidrogenasie
                                                                         (1)
4.2                                                            
        H                   H                          H       H        H
                                +    
   H C          C       C           H O H            H C       C        C H
                      
        H       H           H                          H                H
                                                               O
                                                           H
                                                                                  (4)

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Physical Science/P2                        4                     LimpopoDoE/September 2014
                                     NCS-Memorandum
4.3 Propan-2-ol /2-propanol                                                            (1)

4.4.1 Substitution reaction/ substitusie                                                (1)

4.4.2 Warm (heat) dilute NaOH or KOH (under reflux)/ hitte, verdunde NaOH (2)

4.5.1     Polypropene        / polipropeen                                             (1)

4.5.2     Addition  /addisie                                                           (1)

                                                                                       [11]

QUESTION/VRAAG 5

5.1     CO2 gas generated escapes to the air / CO2 gas vrygestel ontsnap in die lug
                                                                                (1)
                                     ∆m
5.2.    Average rate / gem tempo=
                                      ∆t
                                  4,4 − 0
                                =
                                  240 − 0 
                           = 0,0183 g·s-1                                      (3)
                                           
5.3     4,4 g                                                                  (2)

5.4     All the calcium carbonate has reacted / al die CaCO3 het gereageer
             No more CO2 is being generated / geen CO2 word meer vrygestel nie
             The reaction has ceased (stopped) / reaksie het opgehou       (1)
                                         m
5.5     n(CO2) produced/geproduseer =       
                                         M
                                  4,4
                                =     
                                  44
                               = 0,1 mol
        From the balanced equation/ vanaf gebalanseerde vergelyking:
                 n(CaCO3)      = n(CO2)
                               = 0,1 mol

            m(CaCO3)            = nM
                                = (0,1)(100)
                                = 10 g                                                 (6)
                                                                                       [13]
QUESTION/VRAAG 6

6.1     The stage in a reversible reaction at which the rate at which A(g) and B(g) react
        to form C(g) equals the rate at which C(g) decomposes to reform A(g) and
        B(g)// Dit is 'n dinamiese ewewig wanneer die tempo van die voorwaartse
        reaksie gelyk is aan die tempo van die terugwaartse reaksie.
        OR:
        The stage in a reversible reaction at which the concentrations of A(g), B(g)
        (reactants) and C(g) (products) remain constant.// die stadium in ‘n omkeerbare
        reaksie wanneer die konsentrasie van reagense en produkte constant bly (2)
6.2       H < 0 / H is negative / H is negatief                                        (1)

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Physical Science/P2                            5                    LimpopoDoE/September 2014
                                         NCS-Memorandum
6.3    Decrease the temperature/ verlaag temperatuur
       The forward reaction is exothermic.
       According to Le Chatelier’s Principle, a decrease in temperature places a stress
       on the system. 
       To counteract the stress, the system reacts by favouring the exothermic
       reaction/ the reaction that releases heat.The forward reaction is favoured /
       Equilibrium position shifts to the right/Product formation is favoured.
       Hence the concentration of C (g) increases. 
       Volgens Le Chatelier se beginsel word ‘n stress op die sisteem geplaas
       wanneer die temperatuur verlaag word. Die eksotermiese reaksie word dan
       bevoordeel, d.w.s die voorwaartse reaksie. Dus word meer produk gevorm en
       konsentrasie C neem toe                                                    (4)

6.4    CALCULATIONS USING MOLES:
       Mark allocation:

                                     N                                    N
       •    USING formula n =                / gebruik formule n(A) =
                                       NA                                       NA
       •    Using ratio A(g) : B(g) : C(g) = 2 : 3 : 2/ gebruik verhouding A : B : C = 2 : 3 : 2
       •    Equilibrium n(A) = initial – change
            Equilibrium n(B) = initial – change                 
            Equilibrium n(C) = initial + change
            Ewewig n(A) = aanvanklik - verandering
            Ewewig n(B) = aanvanklik - verandering                     
            Ewewig n(C) = aanvanklik + verandering 
       •    Divide n(A), n(B) & n(C) by 1 dm3 /deel deur 1 dm3
       •    Correct Kc expression (formulae in [ ])/ korrekte Kc uitdrukking
       •    Substitution of concentrations into Kc expression vervang konsentrasies
       •    Final answer/ finale antwoord : 0,064
            OPTION / OPSIE 1

                      N  2,41× 1023
            n(A) =    =             = 0,4 mol
                   NA    6,02 × 1023

            R                  Ag(g)         B(g)            C(g)

            Initial (mole)     2             4               0
            aanvanklik
            Change/ver-        -0,4          -0,6            +0,4          ratio
            andering
            Equ. Mol           1,6           3,4             0,4    
                                                       
            [Eq]               1,6           3,4             0,4           (div)

       Kc =
              [C]
                   2

                       
            [A ]2 [B]3

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Physical Science/P2 6 LimpopoDoE/September 2014
NCS-Memorandum

__ lo4F
~ fyeF 34h

= 0,00159¥ (1,59 x 10°) (8)
OR/of:
CALCULATIONS USING CONCENTRATIONS /berekening met konsentrasies

Mark allocation:

° Using the formula n(A) = v / gebruik formule n(A) ~
A A

Divide n(A) & n(B) by 1 dm°v/ deel deur 1 dm?°

USING ratio: A: B: C = 2:3: 2v/ gebruik verhouding A:B:C =2:3:2
° Equilibrium n(A) = initial — change }

Equilibrium n(B) = initial — change v

Equilibrium n(C) = initial + changev

Ewewig n(A) = aanvanklik - verandering Ewewig

n(B) = aanvanklik - verandering v

Ewewig n(C) = aanvanklik + verandering

Correct K; expression (formulae in [ ])“/ korrekte Ke, uitdrukking

Substitution of concentrations into K, expression” /vervang konsentrasies
. Final answer /finale antwoord: 0,064v

OPTION /opsie 2

[A]= n/v = 2/1 = 2 mol-dm®
[B]= 4/1 = 4 mol: dm®

N ,_ 241x 10%

A) =— Vv = ———_,y;, = 0,4
n(A) Nh 60210" 0,4 mol
A(g) Big) Cig)
R 2 3 2
| (mol/dm*) 2 4 0
C (mol/dm*) -0,4 -0,6 +0,4Y _| ratio/verhouding
E(mol/dm*) a6 | 34%/ 04v
[] equ 1,6 3,4 0,4 v +1
c=", (mol-drv*)
K, = [cf v
[AF ier
_
~ (2@.5)3
= 0,064Y (8)
6.5 Remains the samev / bly dieselfde (1)

[16]

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Physical Science/P2 7 LimpopoDoE/September 2014

NCS-Memorandum

QUESTION/VRAAG 7

7A

7.2
7.2.1

7.2.2

7.3.

7A.

A solution that contains a low (number of mole) amount of acid in proportion to
the volume of water.v v//‘n oplossing wat min (aantal mol) suur in verhouding
tot die volume water bevat (2)

M(KOH) = 39 + 16 + 1 = 56 g:mol"’
m(KOH) = cMVY = (0,2)(56)v (0,3)Y = 3,36 gv

OR/ OF
n=c-V = (0,2)(0,3) = 0,06 mol
m = nM = (0,06)(56) = 3,36 g (4)

KOH “28k* 4 OH
0,2 mol KOH yield 0,2 mol OH”

« [OH] = [KOH]
= 0,2 mol-dm?v
Kw =[HsO*][OH] = 1 x 10%4v

« [HsO*]=1 x 107'4/0,2
=5x 10"4 mol-dm?’v
pH = -log[H,O*]Y”
=-log (5x 10*)v
= -(-13,3)
= 13,3v
OR/ OF:
pOH =-log [OH]Y¥
= - log (0,2)v
=0,7¥
But pH + pOH = 14v
«pH =14—pOH
=14-0,7¥
= 13,3v (6)

Bromothymol bluev/ broomtimolblou
H2SOz is a strong acid and KOH is a strong basev. Therefore the equivalent
point will be at approximately pH=7¥ //H2SO, is ‘n sterk suur en KOH is ‘n sterk
basisv .Daarom is die ekwivalentpunt omtrent by pH = 7 (3)
CaVa _ May
CoVb Mb y
_ C,(20,0) 1 :
“ 2)050y =5 * (ratio)
(0,2)(15,0)

= SE! = 0,075 mol-dm?v 5
a=" (2)(2000) more °)

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Physical Science/P2                          8                    LimpopoDoE/September 2014
                                       NCS-Memorandum
        OR:

        PaVaCa = PbVbCb
         (2)(20,0)Ca=(1)(15,0)(0,2)                  (ratio)
                (0,2)(15,0)(0,2)
         Ca =
                    (2)(20)
              = 0,075 mol·dm-3

        OR/ OF
          n(KOH) = cV
                  = (0,2)(15,0x 10-3)
                  = 3 x 10-3 mol
        But from the balanced equation:/ vanaf gebalanseerde vergelyking
        n(H2SO4)=½ x n(KOH)
                  = ½(3x10-3)
                  = 1,5 x 10-3 mol
          [H2SO4] = n/V
                        1,5 × 10−3
                     =             
                       20,0 × 10−3
                     = 0,075 mol·dm-3                                                   (5)

                                                                                        [20]
QUESTION/VRAAG 8

8.1     Salt bridge / soutbrug
        It maintains electrical neutrality in the cell/It completes the circuit
        (ANY ONE)
        Dit handhaaf elektriese neutraliteit/ dit voltooi die stroombaan (enige een) (2)

8.2     The reducing ability of zinc is greater than that of chromium,             which
          is in turn greater than that of silver. // die reduseervermoë van sink is
          beter as die van chroom en dit is weer beter as silwer s’n          (2)

8.3     Voltmeter’s terminals have been connected incorrectly
          (+ to anode and – to cathode) // die voltmeter se terminale is verkeerd om
          gekoppel (+ aan die anode en - aan die katode)                    (2)

8.4     Zinc is a better reducing agent than chromium and chromium is a better
           reducing agent than silver. / Sink is ‘n beter/sterker reduseermiddel as
           chroom en chroom is ‘n beter reduseermiddel as silwer                  (2)
8.5.1
        Ecell = Ecathode - Eanode    / Esel = Ekatode - Eanode

              = -0,74 – (-0,76) 

        Ecell = + 0,02 V

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Physical Science/P2 9 LimpopoDoE/September 2014
NCS-Memorandum
OR: Cr** +3e° > Cr f E'= -0,74v
Zn >Zn** 426 Jv E* = 40,76”
E®..= 0,02Ww
(4)
8.5.2. Yesv /Ja
Ece > 0 / Ecen is positive / Ege: is positief (2)
8.5.3 Decreases vv /afneem (2)

8.6 Measurements not done under standard conditions of temperature of 25°CVY
and concentration of electrolytes of 1 mol:dm°v
(Accept: internal resistance from the salt bridge) //
Lesing nie onder standaard toestande van temperatuur 25°C en konsentrasie
van elektroliete 1 mol-dm™® geneem (Aanvaar interne weerstand van soutbrug)

(2)

[19]
QUESTION/VRAAG 9
9.1 BY (1)
9.2 Mass deposited is directly proportional to currentv/ massa neergeslaan is
direk eweredigg aan die stroom (1)
9.3
9.3.1 The same asv /dieselfde as (1)
m
=—Vv
9.3.2 n(Cu) M
32
= = v
63,5 0,504 mol Cu (2)

9.3.3 32,0 g Cu = 0,504 moles of Cu/ mol Cu
32 g CU — 0,504 mol electrons/elektrone

-. 63,5 g Cu — 2 mol electrons/ elektrone

Q =ne = (1,008 x 6,02 x 107%)(1,6 x 10°)”
=9707C
But/maar Q. =I-Atv
9707 = 1-(3600)”

“1 =2,70 AY (5)
[10]

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Physical Science/P2                      10                  LimpopoDoE/September 2014
                                   NCS-Memorandum
QUESTION/VRAAG 10
10. 1 By the fractional distillation of liquid air /fraksionale distillasie van lug (1)
10.2
10.2.1 Ammonia/ NH3        / ammoniak                                               (1)
10.2.2 Sulphur/ S /swael                                                            (1)
10.2.3 Ammonium sulphate/ (NH4)2SO4 / ammoniumsulfaat
                    (1)
10.3 Vanadium pentoxide/ Divanadium pentoxide/ V2O5 / vanadiumpentoksied (1)
10.4 Nitrogen  / stikstof                                                           (1)
10.5 2SO2(g) + O2(g)       2SO3(g)
            (Reactants  Products Balancing)                                       (3)

10.6   (ANY ONE)
          • Soil cannot replenish nutrients at a fast enough rate to sustain plant
             growth. / grond kan nie vining genoeg voedingstowwe vrystel om
             groei van plante te onderhou
          • Globally a bigger demand for food. .  /groter wêreldwye behoefde
             aan kos                                                               (1)

                                                                                    [10]

                                               GRAND TOTAL/ Groot Totaal: 150

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